Question: Simplify and expand the following expression: $ \dfrac{2}{a - 9}+ \dfrac{4}{3a - 27}- \dfrac{4a}{a^2 - 18a + 81} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the second term: $ \dfrac{4}{3a - 27} = \dfrac{4}{3(a - 9)}$ We can factor the quadratic in the third term: $ \dfrac{4a}{a^2 - 18a + 81} = \dfrac{4a}{(a - 9)(a - 9)}$ Now we have: $ \dfrac{2}{a - 9}+ \dfrac{4}{3(a - 9)}- \dfrac{4a}{(a - 9)(a - 9)} $ The least common multiple of the denominators is: $ (a - 9)(a - 9)$ In order to get the first term over $(a - 9)(a - 9)$ , multiply by $\dfrac{3(a - 9)}{3(a - 9)}$ $ \dfrac{2}{a - 9} \times \dfrac{3(a - 9)}{3(a - 9)} = \dfrac{6(a - 9)}{(a - 9)(a - 9)} $ In order to get the second term over $(a - 9)(a - 9)$ , multiply by $\dfrac{a - 9}{a - 9}$ $ \dfrac{4}{3(a - 9)} \times \dfrac{a - 9}{a - 9} = \dfrac{4(a - 9)}{(a - 9)(a - 9)} $ In order to get the third term over $(a - 9)(a - 9)$ , multiply by $\dfrac{3}{3}$ $ \dfrac{4a}{(a - 9)(a - 9)} \times \dfrac{3}{3} = \dfrac{12a}{(a - 9)(a - 9)} $ Now we have: $ \dfrac{6(a - 9)}{(a - 9)(a - 9)} + \dfrac{4(a - 9)}{(a - 9)(a - 9)} - \dfrac{12a}{(a - 9)(a - 9)} $ $ = \dfrac{ 6(a - 9) + 4(a - 9) - 12a} {(a - 9)(a - 9)} $ Expand: $ = \dfrac{6a - 54 + 4a - 36 - 12a}{3a^2 - 54a + 243} $ $ = \dfrac{-2a - 90}{3a^2 - 54a + 243}$